There was this question, whether a torus in dimension n, $T^n$, can carry a riemannian metric with positive sectional curvature.
A read a proof, which goes as follows:
$T^n$ is complete, because it's compact. Assume $sec_x>0$ for all $x\in T^n$. Then his universal covering has also positive sectional curvature, hence is compact after the Bonnet-Myers theorem. So the fundamental group of $T^n$ is finite, which is a contradiction.
I think this is no complete proof. The Theorem of Bonnet-Myers needs an unifome bound on the curvature, so strictly positive curvature everywhere is not sufficient. Here's mny counterexample: Take the upper hyperboloid in the three dimensional euclidian space $M:=\{x\in\mathbb{R}^3|x_1^2+x_2^2-x_3^2=-1,x_3>0\}$ with the induced metric from $R^3$. This has positive sectional curvature evreywhere, but is not compact. (Because as $|x|$ goes to infinity, the sectional curvature of $x$ goes to zero.)
What do I make wrong? Or am I right and this proof is incomplete? If so, how can I get an uniformly bound on the sectional curvature? Using that $T^n$ is compact? How can I make this precise?
Your hyperboloid has negative sectional curvature everywhere, indeed constant $-1.$ This is one of the standard models of the hyperbolic plane. http://en.wikipedia.org/wiki/Hyperboloid_model
In case this is the source of the confusion, the restriction of the Minkowski form to the hyperboloid becomes a positive metric, that is, Riemannian.